tan(1/2(cos^−1(x))=√1−x/1+x

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tan(1/2(cos^−1(x))=√1−x/1+x

Trigonometry
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Do you have any idea on starting this question?
no is say prove that their equal
Well that's the first step :) Do you know what to do next?

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Other answers:

no
We don't need to rush in and do everything in one go :) We can firstly start of with: tan(1/2(cos^−1(x))
Tan ( 1 ) 2 Cos^-1(x)
Try changing the inverse trig ratio
You get something like: Tan (Secx) 2
ok
So graphically, it looks something like this
|dw:1433678743148:dw|
The only way to get the prove this, is by using properties of Pythagoras's, since we are dealing with a right angle triangle
ok so we finding the hypotenuse in other word?
Yeah we could try to find the hypotenuse or the other side. Just some way to get us that Square root, to help us prove that LHS = RHS
so we would have x^2=(1/cosx)^2+2^2)?
Question: \(tan (\dfrac{1}{2} arccos (x))\) or \(tan (\dfrac{1}{2arccos (x))}\). which one?
tan(12arccos(x))
the first 1
if so, then the right hand side must be \(\dfrac{\sqrt {1-x^2}}{1+x}\) , right?
you missed the square at x from numerator. Am I right? please check the problem
|dw:1433681866420:dw|
the right side has no square at x its just; the root of (1-x)/(1+x)
If it is so, I failed. :)

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